Problem 48.Self Power.....The series, 11 + 22 + 33 + ... + 1010 = 10405071317. Find the last ten digits of the series, 11 + 22 + 33 + ... + 10001000.

main()
{
int num;
int j;//to save the value of i
int i;
int reserve;//for reserving the value of number.
static int sum[10];//for storing the value of result and static for making all array value initially zero.
static int a[10];//for storing value of individual power
int n=10;
int check;//for checking wether an element of array is getting carry or not.
for(num=1;num<=1000;num++)//Loop for taking all the numbers whose power is taken.
{
a[n-1]=num;
for(reserve=1;reserve<num;reserve++)//Loop for value of individual Power.
{
for(i=0;i<n;i++)//Loop for multiplying each array element with the number to get power;
{
a[i]*=num;

}

for(i=0;i<n;i++){//Loop for making all element of array single digit
if(a[i]>9)
{
check=a[i];
a[i]=(check%10);
check/=10;
j=i-1;
while(check!=0)
{
if(j<0){break;}
a[j]+=(check%10);
check/=10;
j--;
}
}
}
}
for(i=0;i<n;i++)//Summing the value of individual self power in sum array
{
    sum[i]+=a[i];
a[i]=0;//To reinitialize all value of array to zero for next self power
//Now making all the value of array sum as single digit
    if(sum[i]>9)
    {
            check=sum[i];
            sum[i]=check%10;
            check/=10;
            j=i-1;
            while(check!=0)
            {if(j<0){break;}
                    sum[j]+=(check%10);
                    check/=10;
                    j--;
            }

}

}
}
for(i=0;i<=9;i++)//for printing last 10 digits
{
    printf("%d",sum[i]);
}
}